Is this maths correct?

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Offline jccc

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Re: Is this maths correct?
« Reply #100 on: 05/07/2015 19:18:54 »
I will give up , although I know very well I am correct,
You give up without having read and understood our posts. For that reason you will never understand why you are wrong, and you will never learn maths and probability.
A pity, because you could have.

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/53 is correct for every deck.

100 cards you get 100/13  aces. 

i see your fishing trip is very successful, good job bro!


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Offline Thebox

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Re: Is this maths correct?
« Reply #101 on: 05/07/2015 19:23:58 »
I will give up , although I know very well I am correct,
You give up without having read and understood our posts. For that reason you will never understand why you are wrong, and you will never learn maths and probability.
A pity, because you could have.

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/53 is correct for every deck.

100 cards you get 100/13  aces. 

i see your fishing trip is very successful, good job bro!

7.69230769231≠0.07692307692   

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Offline chiralSPO

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Re: Is this maths correct?
« Reply #102 on: 06/07/2015 02:16:01 »

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/52 is correct for every deck.

100 cards you get 100/13  aces. 

i see your fishing trip is very successful, good job bro!

jccc is absolutely correct here.


7.69230769231≠0.07692307692   


No, it is exactly 100 times bigger (because you're drawing 100 cards, not 1)...

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Offline jccc

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Re: Is this maths correct?
« Reply #103 on: 06/07/2015 02:29:19 »

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/52 is correct for every deck.

100 cards you get 100/13  aces. 

i see your fishing trip is very successful, good job bro!

jccc is absolutely correct here.


please save that line for my gravity/light theory.

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Offline Thebox

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Re: Is this maths correct?
« Reply #104 on: 06/07/2015 06:15:31 »

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/52 is correct for every deck.

100 cards you get 100/13  aces. 

i see your fishing trip is very successful, good job bro!

jccc is absolutely correct here.


7.69230769231≠0.07692307692   


No, it is exactly 100 times bigger (because you're drawing 100 cards, not 1)...

No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.

x is not equal to  y. I believe I am correct in my idea?

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Offline Colin2B

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Re: Is this maths correct?
« Reply #105 on: 06/07/2015 09:18:31 »
jccc is right
Consider y
Toss a coin. 1/2 chance for any face value H or T. Toss 100 times still 1/2 each throw. Toss  ∞ times  still 1/2 each throw. yes you can get multiple heads in 100 throws but they even out over a large number of throws so in 100 throws you are more likely to get 50 H and 50 T.

Now throw a dice. 1/6 chance of any face value. Throw 100 times still 1/6 each throw.  Throw ∞ times still 1/6 each throw. Again evens out over a large number of throws.

Now throw a 52 sided dice with a card value painted on each face. Better still deal a shuffled pack and choose the top card. 1/52.
As with the coin and 6 sided dice you can get repeats but they even out. The probabilities or odds are not unknown.
Probability of repeats?
4 H in a row 1/2*1/2*1/2*1/2=(1/2)4
4 ace of diamonds in a row (1/52)4= not very likely.
100 coin tosses you are more likely to get 50 H and 50 T.
Draw top card from 100 decks, then you are likely to have 100/13 aces.
In reality, you will get some slight variations from these, but over a large number eg >300 the actual numbers get closer to the calculated probabilities.
These are not unknown probabilities, they are known and understood.

Happy fishing by the way.

PS good one jccc you understand probability, we'll have you understanding QM soon. Probability of electron etc  [;)]


EDIT:

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
.
.
No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.
These are not the same question. On the 1st jccc is right.
On the 2nd read what ChiralSPO and I have written above then consider:
If you toss a coin 99 times what is the probability of a H on the next toss? it is 1/2.
Now get someone to toss a coin 100 times and write down the outcomes. Then you choose one of them, if you choose #100 then the probability it is a H is still 1/2. You are not choosing from 100, you are choosing from 2 possibilities, either it is a H or it is a T = 1/2.
So if you choose a deck from 100 preshuffled decks the probability of an ace top card is 4/52. The 99 other decks are irrelevant, you are only choosing this one.

I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.


« Last Edit: 06/07/2015 11:21:04 by Colin2B »
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #106 on: 06/07/2015 15:52:28 »
jccc is right
Consider y
Toss a coin. 1/2 chance for any face value H or T. Toss 100 times still 1/2 each throw. Toss  ∞ times  still 1/2 each throw. yes you can get multiple heads in 100 throws but they even out over a large number of throws so in 100 throws you are more likely to get 50 H and 50 T.

Now throw a dice. 1/6 chance of any face value. Throw 100 times still 1/6 each throw.  Throw ∞ times still 1/6 each throw. Again evens out over a large number of throws.

Now throw a 52 sided dice with a card value painted on each face. Better still deal a shuffled pack and choose the top card. 1/52.
As with the coin and 6 sided dice you can get repeats but they even out. The probabilities or odds are not unknown.
Probability of repeats?
4 H in a row 1/2*1/2*1/2*1/2=(1/2)4
4 ace of diamonds in a row (1/52)4= not very likely.
100 coin tosses you are more likely to get 50 H and 50 T.
Draw top card from 100 decks, then you are likely to have 100/13 aces.
In reality, you will get some slight variations from these, but over a large number eg >300 the actual numbers get closer to the calculated probabilities.
These are not unknown probabilities, they are known and understood.

Happy fishing by the way.

PS good one jccc you understand probability, we'll have you understanding QM soon. Probability of electron etc  [;)]


EDIT:

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
.
.
No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.
These are not the same question. On the 1st jccc is right.
On the 2nd read what ChiralSPO and I have written above then consider:
If you toss a coin 99 times what is the probability of a H on the next toss? it is 1/2.
Now get someone to toss a coin 100 times and write down the outcomes. Then you choose one of them, if you choose #100 then the probability it is a H is still 1/2. You are not choosing from 100, you are choosing from 2 possibilities, either it is a H or it is a T = 1/2.
So if you choose a deck from 100 preshuffled decks the probability of an ace top card is 4/52. The 99 other decks are irrelevant, you are only choosing this one.

I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.


I am not making no mistakes, my logic is accurate, but from this post I now know why you are looking always from the wrong angle.
You are not considering that the coin is already tossed 100 times.

You are not considering the already set unknown sequence of a deck of cards.

The top card is not random, it is a unknown value.  Once the top card is set, it is set and nothing can change this.

You are making a mistake and still looking a the wrong angle and perspective.


Answer this,

Take two decks of cards for simplicity, try it at home, shuffle both decks individually,

both decks have a probability that an ace is the top card of 4/52.

I ask you to take the top card of each deck , and discard the other cards of the decks, at this stage there is still a 4/52 chance that any of the two cards is an ace.

I have a quick look, and can confirm that one of the cards is an ace. 

Please state the probability of the two cards now?

added -
100*4/52 = 7.69, which is roughly 8.

a
x
x
x
x
a
x
x
x
x
a
a
a
x
x
x

quantum leap , skip some space time, get lucky and land on another a.

it is very simple

pick a card anywhere from in a deck. 1/52

pick a card from another deck, 1/52


and continue this 100 times.

x
x
x
x
x
x
x
x
x
and so on to 100.


now what is the probability of drawing an ace from the new made y axis?





You are player 1, you will get card 1, pick a deck

1-52
1-52
1-52
1-52
and so 100 times in total.

You are not picking from 1-52, you are picking from 1-100 of unknown variant quantities.



 






« Last Edit: 06/07/2015 20:06:32 by Thebox »

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Offline Colin2B

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Re: Is this maths correct?
« Reply #107 on: 07/07/2015 00:07:46 »

You are not considering that the coin is already tossed 100 times.
It doesn't matter. Previous tosses do not affect future tosses. This is a common error, the coin has no memory. Probability is 1/2 every time

You are not considering the already set unknown sequence of a deck of cards.
Probability isn't about a specific set sequence, it is the likelihood of an event occurring or having occurred and is based on lots of trials not just the one or two you have carefully selected to make your point.

The top card is not random, it is a unknown value. 
A card can be both random and unknown, in fact it usually is both until looked at. So I don't understand what you mean by this.

Once the top card is set, it is set and nothing can change this.
No one is saying you can change it, but probability only deals with what is likely or unlikely not with what is actually there.

You are not picking from 1-52, you are picking from 1-100 of unknown variant quantities.
This is wrong as already explained.

I ask you to take the top card of each deck , and discard the other cards of the decks, at this stage there is still a 4/52 chance that any of the two cards is an ace.
No the probability is not 4/52, ChiralSPO and I explained this in posts #54 & 55.
You clearly haven't read these posts. I'm not going to repeat them again, please read them.

I have a quick look, and can confirm that one of the cards is an ace. 

Please state the probability of the two cards now
Again, this was explained fully in those 2 posts. This is condition probability which I have also explained in other posts. It is a mistake to assume P(AB)=P(B|A) as I previously explained.

If you are not going to bother reading and understanding our posts there really isn't any point continuing. I have persisted this far because I am aware that in a few years your son will be starting secondary school. His first exposure to probability in maths will be through these very basic first year examples using coins and cards. If you teach him your pseudomaths he will fail his coursework and be considered a fool by his friends and teachers. This will be unfair. You owe it to him to learn real maths, not myth.

Go back, read what Alan, ChiralSPO and I have written. If you have specific questions on understanding what we have written I will try to answer, but I am unable to deal with your delusion regarding the 'timing' of decks and it's effect on outcomes.

Whatever it is you think you are doing it isn't probability.

Like Alan, I'm out.

and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #108 on: 07/07/2015 15:50:19 »
We are in just chat , chatting, out of what exactly?

And like every other forum on the internet, again your versions are being forced onto me. The things you defend are not even your things, they were some ones from history things.

I ask where is your own thought and consideration for the talking point?  without resorting back to present information.
Does nobody on this planet have the ability to think about what is actually said?
I wish to discuss x is not equal to y, I have shown axiom logic and models and maths. All you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point. 
You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.


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Offline Colin2B

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Re: Is this maths correct?
« Reply #109 on: 07/07/2015 17:53:28 »
I ask where is your own thought and consideration for the talking point? 
These are my own thoughts.
I have looked at probability, I have done the experiments and the maths and understood.

I wish to discuss x is not equal to y, I have shown axiom logic and models and maths.
And we have shown by logic, models and maths that we disagree with you

All you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point.
If you tell me there are no fish in the sea, I have to disagree with you.
If you tell me there is no ocean between UK and US, I have to disagree with you.

I have to be true to myself, I cannot agree with false 'facts',  false logic or false maths.

You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.
No, I can visualise what you are trying to say and I understand it, but I also understand why you are seeing it incorrectly.
Again I cannot agree with falsehoods and it would not be respectful to you if I did.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #110 on: 07/07/2015 19:01:05 »
I ask where is your own thought and consideration for the talking point? 
These are my own thoughts.
I have looked at probability, I have done the experiments and the maths and understood.

I wish to discuss x is not equal to y, I have shown axiom logic and models and maths.
And we have shown by logic, models and maths that we disagree with you

All you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point.
If you tell me there are no fish in the sea, I have to disagree with you.
If you tell me there is no ocean between UK and US, I have to disagree with you.

I have to be true to myself, I cannot agree with false 'facts',  false logic or false maths.

You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.
No, I can visualise what you are trying to say and I understand it, but I also understand why you are seeing it incorrectly.
Again I cannot agree with falsehoods and it would not be respectful to you if I did.

I am not telling you that there is no fish in the sea, I am telling you that there is too may fish in the sea.

100 decks of randomly shuffled cards will contain approx 8 aces as the top card of the 100 decks.


The small blind is in alignment with card 1 of each deck. 

''There are 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 arrangements of a set of 52 cards. 7.69% of them have an ace at the top.''

So if I have 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000  decks of cards pre-shuffled, column 1 to the small blind contains 7.69% worth of aces.  compared to 4/52

it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.
« Last Edit: 07/07/2015 19:02:51 by Thebox »

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Offline Colin2B

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Re: Is this maths correct?
« Reply #111 on: 08/07/2015 04:39:03 »
the small blind contains 7.69% worth of aces.  compared to 4/52
My dear Box, at last we are in agreement.
Yes 7.69% does indeed = 4/52
Now you can see that the probability in the y direction is the same as in the x direction.
P(y)=P(x)
y=x

Now at last you understand what we are saying and we can begin to explore the ways in which x and y are different.

it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.
Yes, I pointed this out to you many posts ago. However, as I also pointed out it is extremely unlikely to happen in anybody's lifetime.

There has never been any question that there are different sequences in the x and y directions, the problem has been your misinterpretation of the probabilities and the effect they have on the games you play. You asked "is this maths correct" and we have explained that it is not correct.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #112 on: 08/07/2015 16:31:24 »
the small blind contains 7.69% worth of aces.  compared to 4/52
My dear Box, at last we are in agreement.
Yes 7.69% does indeed = 4/52
Now you can see that the probability in the y direction is the same as in the x direction.
P(y)=P(x)
y=x

Now at last you understand what we are saying and we can begin to explore the ways in which x and y are different.

it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.
Yes, I pointed this out to you many posts ago. However, as I also pointed out it is extremely unlikely to happen in anybody's lifetime.

There has never been any question that there are different sequences in the x and y directions, the problem has been your misinterpretation of the probabilities and the effect they have on the games you play. You asked "is this maths correct" and we have explained that it is not correct.

Thank you Colin, interesting that probabilities may not be the answer I am looking for, although alternative scenarios must apply?

I do not why you think I am agreeing with the maths and that Y=X.  Not for one second do I think Y=X.


Forget probabilities for a while and please answer me this, in the diagram below , does Y=X?

.
.
.
.  .  .  .






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Offline alancalverd

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Re: Is this maths correct?
« Reply #113 on: 08/07/2015 17:25:51 »
I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.

Good point! In fact if you have just tossed 5H, it's quite likely that the coin is biassed so the next toss is more likely to be H than T!
helping to stem the tide of ignorance

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Offline Thebox

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Re: Is this maths correct?
« Reply #114 on: 08/07/2015 17:31:13 »
I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.

Good point! In fact if you have just tossed 5H, it's quite likely that the coin is biassed so the next toss is more likely to be H than T!


^5  reverse bet not fallacy, I would bet you could not throw another heads, I am not betting 1/2 of heads or tails



and I forgot science likes simple


x
x
x
a
b
c


x=1 or 2


x~abc random over time.

what is the chance that (a) will receive a 1.
« Last Edit: 08/07/2015 17:33:02 by Thebox »

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Offline alancalverd

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Re: Is this maths correct?
« Reply #115 on: 09/07/2015 00:04:51 »
, I would bet you could not throw another heads,

Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.
helping to stem the tide of ignorance

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Offline Thebox

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Re: Is this maths correct?
« Reply #116 on: 11/07/2015 09:33:52 »
, I would bet you could not throw another heads,

Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.

You have not got me, I do not ignore solid evidence, I know the coin still has 1/2 chance.  And this has nothing to do with my poker ideas.
I understand randomness, I understand sequence, and I certainly understand x and y. It is not my misconception, it is the entire human races perspective view of just about everything,  from atoms to probability.

The human race looks at things in a 1 dimensional perspective, where everything in my space is see through.

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Offline Thebox

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Re: Is this maths correct?
« Reply #117 on: 11/07/2015 10:30:18 »
I refuse to back down from this when I know I am correct and in this post I will concentrate on what I am writing and try to get through an understanding of mine that science seems to be missing.

It has come to my attention by playing online poker, and by my observations and knowledge of the distribution process that there is a flaw in the process. The flaw is a probability function, that at the start of any hand, all players have  uneven probabilities, a flaw caused by a multitude of pre-shuffled decks, that are randomly issued to game tables each and every hand.
A time based distribution, based on when a table finishes it's hand, it gets a new deck from the system that is pre-waiting in storage.
This makes an anomaly, where  standard random sequences over time from using a single deck, is altered by periodic ''jumping'' of spacial dividers.

We can assert that of a multitude of individual pre-shuffled decks, that  the  top cards of each deck will all have a 4/52 chance of it being an ace, we can with a certainty be assured that a single deck only contains 4 aces. However we can also also assert that the top card of each deck could all be an ace, or all could be a different card to an ace.

If we consider a diagram form and a variation of top cards,

ace
queen
four
ace


the four cards representing 4 individual decks and the top card of each deck, if a table received deck 1, being at the top , and deck 4 being the at the bottom, then that table card 1 is both times an ace.  This is the spacial dividers over time ''jumping'' I refer to.


ace,queen,four,ace

ace........time........ace

it is not hard to consider an oblong format of random sequences of rows,

Obviously x axis does not equal y axis,

123
231
123


X makes y, and Y is different to x, x remains 1/3 whilst Y remains unknown ,

I am not saying ''that the probability of getting an ace decreases as the number of decks increases''.


I am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.

If we call 1, (n), P(n)/x=1/3 but P(n)/y=?, because we could never possibly know what values had fell in place of the Y axis columns, example

123
123
123

In this situation column one on the left, the P(1)/y=3/3 where as x always remains 1/3.
And if we hide the values, the result could still be the same.

nnn
nnn
nnn

we know left to right of each row is still 1/3, but we do not know the columns values.











« Last Edit: 11/07/2015 10:35:23 by Thebox »

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Offline alancalverd

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Re: Is this maths correct?
« Reply #118 on: 11/07/2015 19:45:03 »

Quote
I am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.

This is entirely due to the randomness of the distribution of aces. It is known as the luck of the draw, or the consequence of shuffling, and is what turns poker from an algorithm into a game.

However your misappropriation of the term "quantum leap" is deplorable in a science forum. What you are saying is that if you sit at a random table, there is a 1/13 chance that the first card you receive will be an ace, and the probability of it happening twice in two successive independent shuffles is 1/169.
helping to stem the tide of ignorance

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Offline Thebox

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Re: Is this maths correct?
« Reply #119 on: 11/07/2015 22:18:50 »

Quote
I am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.

This is entirely due to the randomness of the distribution of aces. It is known as the luck of the draw, or the consequence of shuffling, and is what turns poker from an algorithm into a game.

However your misappropriation of the term "quantum leap" is deplorable in a science forum. What you are saying is that if you sit at a random table, there is a 1/13 chance that the first card you receive will be an ace, and the probability of it happening twice in two successive independent shuffles is 1/169.

I do not see how advancing forward is deplorable in a science forum?

I have no idea where you have 1/169 from?

or 1/13?

I have one deck of cards 4/52 chance of an ace following my x axis, you you can make the x a Y if you like and the result is the same.


I have the y axis, at  ?/? .

12
21


x=y


21
21

x≠y

only 2 values are needed to show this.

12
21

xy≠xy
« Last Edit: 11/07/2015 22:21:59 by Thebox »

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Offline alancalverd

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Re: Is this maths correct?
« Reply #120 on: 12/07/2015 00:46:52 »
4/52 = 1/13

If P = probability of something happening in one trial, the probability of it happening n times in succession in independent trials = P^n.

So probability of the first card being an ace in two (and only two) independent trials = (1/13)^2 = 1/169

A posteriori analysis of a small number of trials does not give you the probability of an event. Choosing or constructing two or three possible sequences tells you nothing about the likely outcome of any actual trial, let alone the probability of a given outcome in sequential trials.

For the very last time: if the shuffles are independent (a) the probability of an outcome is exactly the same in each shuffle, by definition of independent, and (b) the statistics of actual outcomes only tend towards the calculated probability for a very large number of trials, by definition of probability.

Belief in anything else is the first step on the road to gambling bankruptcy.
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Offline Thebox

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Re: Is this maths correct?
« Reply #121 on: 12/07/2015 10:41:10 »
4/52 = 1/13

If P = probability of something happening in one trial, the probability of it happening n times in succession in independent trials = P^n.

So probability of the first card being an ace in two (and only two) independent trials = (1/13)^2 = 1/169

A posteriori analysis of a small number of trials does not give you the probability of an event. Choosing or constructing two or three possible sequences tells you nothing about the likely outcome of any actual trial, let alone the probability of a given outcome in sequential trials.

For the very last time: if the shuffles are independent (a) the probability of an outcome is exactly the same in each shuffle, by definition of independent, and (b) the statistics of actual outcomes only tend towards the calculated probability for a very large number of trials, by definition of probability.

Belief in anything else is the first step on the road to gambling bankruptcy.

I know Alan you still do not understand my argument and what I am arguing about.  My argument is not about the probabilities of independent decks.

I know each deck has a 4/52 chance of the top card being , but I  also know this only applies using an x-axis scenario.

To say Y is equal to X when talking probability is absurd when it is evidentially not, no matter how many digits is in use.

xxxxx
xxxxx
xxxxx
xxxxx
xxxxx

5², each row as the values of 1 to 5, these values have been displaced of any order by a random shuffle, when the shuffle stops, in alignment with a Y axis and of 5 columns, each column also has 5 values, can you tell me what values are in the columns?

I can tell you that in every row there is 1 to 5 in no order.




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Offline Colin2B

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Re: Is this maths correct?
« Reply #122 on: 12/07/2015 17:38:01 »
I do not why you think I am agreeing with the maths
Your interpretation of the probabilities in the x&y directions has always been a sticking point in discussing you theory. In particular you think that the probability of an ace when picking from the top cards of 100 decks is different to that of picking from a single deck of 52. Well, in one post you said:
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.
However, later when talking about 100 decks you say:
..........column 1 to the small blind contains 7.69% worth of aces.  compared to 4/52
So, because 7.69%=4/52=8/104=0.769 you are agreeing that probability in columns = probability in rows

I forgot science likes simple
No, science is comfortable with complex. What it does like is clarity. If you ask vague, woolly, unclear or confusing questions you can expect to be asked "what do you mean by...", "define ....", "what are you really asking". You have to be clear, something you are usually not.
However, because you have been much clearer in post #117, I will make one last attempt to show why the probability in x & y are the same, but the sequences are not.

nnn
nnn
nnn

we know left to right of each row is still 1/3, but we do not know the columns values.
We have previously pointed out that the probability of an ace in the y direction is the same as in the x see posts.#54, 55 and 105. What differs, as we have explained, are the sequences.
However, these are not unknown because probability helps us to understand the unknown.

OK, let's take your example as it is fairly simple.

nnn
nnn
nnn

You have 3 decks y1, y2, y3. Rather than use numbers for the cards I am going to use letters because otherwise we get confused when talking about 1st cards, 2nd card etc. So the 3 cards are A, B, C = Ace, Bravo, Charlie
So in the x direction P(x1)=P(x2)=P(x3)=1/3 and we agree on that.
So if you shuffle a deck of these cards and pick the first there is a 1/3 chance it is A, but if you had picked the bottom card that would also be probability of 1/3, same with the middle card.
So in the x direction we are agreed 1/3 1/3 1/3.
So what about y.
If we shuffle your 3 decks and take the top 3 cards y1 y2 y3 then there are 3 possibilities for each y and so the number of possible ways these cards can fall = 3*3*3=27 possibilities.
Let's look at the probability that in these 3 cards the 1st card only is A.
We can show this with maths using an X to show “not an A” = a 2 or a 3
So, AXX
here the probability of A is 1/3, but there are 2 possible cards in y2, a B or a C so the probability is 2/3, same with y3. So total probability = 1/3*2/3*2/3= 4/27.
We can  also do this by writing out the possibilities for this deal:
A in y1
C   B   B   C
B   C   B   C
A   A   A   A
a total of 4 ways out of 27 =4/27

We can also look at the probability that one of the cards out of the 3 is an A regardless of position. To do this we need to look at  Ain y2 and A in y3

A in y2
We can also do this for y2 = A
C    B    B    C
A    A    A    A
B    C    B    C
Another 4 ways

A in y3
and for y3 = A
A    A    A    A
C    B    B    C
B    C    B    C
yet another 4 ways
Total 12

So if we want to know the probability of a single A anywhere in the 3 cards it is = 12/27=4/9

We would also want to look at the other ways for completeness:
A    B    C
A    B    C
A    B    C
3 more ways
and the ones we missed:
A   C   B   C
A   C   C   B
B   B   B   B

A   A   B   C
B   C   A   A
A   A   A   A

B   A   B   C
B   A   C   B
C   C   C   C
Total 27

In order to do this many ways we have had to lay out 3x27 decks = 81
so in all those top cards how many were As?
Count them A = 27
So probability of A occurring in y = 27/81 = 1/3
the same as in x.
You will also note that AAA occurs once so probability = 1/27
but so does CBC, and BCC and BBB. All have equal probability.
Only humans decide that A is special. That an ace wins over a 6 or a 5. In reality their probabilities are all the same!

Now, the chances are that you would not get this exact match in exactly 81 decks, but as Alan said if you do enough games (trials) say >1000, then the actuality will get close to these probabilities.

What this also tells us is that for any 3 decks all arrangements are equally likely, so it doesn't matter if you skip decks or do them in sequence, the probability is the same. Out of a 1000 decks you could choose 3 from anywhere and the probability would be the same for each. If you think otherwise, you do not understand what probability is telling us.

You can obviously do the calculation shown above for any number of decks and cards, but the numbers become large, so I'll leave that to you.

Thank you Colin, interesting that probabilities may not be the answer I am looking for,
I would agree. Probability tells us that there is no difference between shuffled decks, so skipping some of the decks has no effect on the outcome of games from a probability point of view.
Yes, there are other scenarios. Your's sounds closer to predestination, or "that one was intended for me" type of luck or fortune. I know something about probability, permutation and sequences, but nothing about predestination, fortune or luck, so I can't help you with that.
« Last Edit: 12/07/2015 17:43:11 by Colin2B »
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Offline alancalverd

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Re: Is this maths correct?
« Reply #123 on: 12/07/2015 18:41:23 »
, I would bet you could not throw another heads,

Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.

You have not got me, I do not ignore solid evidence, I know the coin still has 1/2 chance.


But the evidence to date suggests that it doesn't! The probability of scoring 5H in a row in inependent trials of a fair coin is (1/2)^5, around  3%, so the evidence suggests a 97% probability that the coin is not fair. On that basis it would be foolish to bet against it.

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Offline Thebox

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Re: Is this maths correct?
« Reply #124 on: 13/07/2015 14:52:42 »
Thank you Alan and Colin for your persistence. Your maths is accurate and true if y1,y2 and y3 were all coming to the same table.

Colin calls it pre destined, I call it probabilities over a time period.

This is my whole point and the only part you are not understanding, the part about interception.

y1,y2,y3,y4,y5 where y is deck
 t1,t2,t3,t4,t5  where t is table



randomly distributing y to each table is not the same.

You say you know sequences Colin, then you understand spacing between values, spacing being equivalent to time.


ace.........end.......ace..............end.................ace..end.  end being end of sequence.


Above I show 3 shuffles and 3 decks, Player 1 table one receives on average 1/221 aces/t


divided by time being a key element when considering probabilities and certainly poker.







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Offline Colin2B

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Re: Is this maths correct?
« Reply #125 on: 13/07/2015 17:39:12 »
Thank you Alan and Colin for your persistence. Your maths is accurate and true if y1,y2 and y3 were all coming to the same table.
No, that's incorrect. As I explained in my post, deck skipping (spacing, interception or timing as you call it) does not affect the outcome.
If you have an infinite number of decks and you wish to choose 3 for each table it does not matter which order you distribute them in or by what timing. You could take the 5th, the 10th and the 81st for one table; the 93rd, 4th and 100006th for the 2nd table etc. All are as equal as if you had taken them in order 1 2 3 4 5 6 etc.
If you had understood my post you would realise that what determines the randomness is not the distribution in the y direction, but the random shuffle of the decks. This makes all decks equal from a probability point of view.
Any other belief is not probability. And as I have said before you cannot divide a probability by time.

So let's leave it that I believe in probability and maths, and you believe in something else.


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Offline Thebox

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Re: Is this maths correct?
« Reply #126 on: 14/07/2015 16:54:23 »
Thank you Alan and Colin for your persistence. Your maths is accurate and true if y1,y2 and y3 were all coming to the same table.
No, that's incorrect. As I explained in my post, deck skipping (spacing, interception or timing as you call it) does not affect the outcome.
If you have an infinite number of decks and you wish to choose 3 for each table it does not matter which order you distribute them in or by what timing. You could take the 5th, the 10th and the 81st for one table; the 93rd, 4th and 100006th for the 2nd table etc. All are as equal as if you had taken them in order 1 2 3 4 5 6 etc.
If you had understood my post you would realise that what determines the randomness is not the distribution in the y direction, but the random shuffle of the decks. This makes all decks equal from a probability point of view.
Any other belief is not probability. And as I have said before you cannot divide a probability by time.

So let's leave it that I believe in probability and maths, and you believe in something else.

Again from your wording I see my very own idea,

''You could take the 5th, the 10th and the 81st for one table; the 93rd, 4th and 100006th for the 2nd table etc. All are as equal as if you had taken them in order 1 2 3 4 5 6 etc.''



This is not true,they are not equal.

y1
y2
y1
y2
y1
y2

That is a completely different distribution pattern if you deck skip .


lets change the order,

y1
y1
y1
y2
y2
y2

lets change the angle for you

y1y1y1y2y2y2
t...................t


can you not see the difference to a unified distribution and skipping time distribution?

You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card, what is the chance you will receive a deck that the ace is the second card?


« Last Edit: 14/07/2015 17:00:10 by Thebox »

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Offline Colin2B

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Re: Is this maths correct?
« Reply #127 on: 14/07/2015 18:56:28 »

This is not true,they are not equal.

y1
y2
y1
y2
y1
y2

That is a completely different distribution pattern if you deck skip .


lets change the order,

y1
y1
y1
y2
y2
y2

lets change the angle for you

y1y1y1y2y2y2
t...................t


can you not see the difference to a unified distribution and skipping time distribution?
Not from a probability point of view. They are all the same.
I was beginning to suspect this is where the core of the problem lies, but let me answer your other question first

You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card, what is the chance you will receive a deck that the ace is the second card?
Probability deals with the likelihood of events happening where we do not know the outcome, but it is based on the knowledge we have at the time.

Ok, to be sure I understand your scenario.
I have just received the 1st deck from the stack and received an ace. You are asking the probability that the 2nd card is an ace. I don't know that that is the way the cards have fallen, so probability says the chance of a 2nd ace is 3/51.
If all the decks have been correctly shuffled the chance is still 3/51.
If you tell me before the game that these decks have not been correctly shuffled, and you have prior knowledge of a different proportion existing, then that changes the probability.

However, I don't think that is what you mean. I think you imagine the decks being set out, with their shuffled order and by chance, unknown to all the players, 10% of the decks have aces 2nd card. This is a possible scenario, but if the players do not know this then it does not change the probability of an ace 2nd card.

I will repeat "Probability deals with the likelihood of events happening where we do not know the outcome, but it is based on the knowledge we have at the time"

As I have described in previous posts, we can see the likely outcomes as probability distributions. What you are talking about are actuality distributions - in other words, what really happens.
This is akin to the QM situation where we do not know the position of an electron, say, but we know the probability of it being in a certain area. It is only when it is observed or captured that its location is known.
In the same way, probability tells us the likely values and locations of cards, but it is only when they are played that we know their  true location and value.

If you wish to deal with actuality distributions, you need to record them, probability can tell you whether that was a likely event and in some cases will be able to tell you how likely it is that the deck was fairly shuffled. Beyond that probability in card games does not deal with actuality.
It may be that the 1st 10 decks all have an ace on top, yes if you get the 11th deck then you have missed out. Probability still says all the decks are equal and before the game was played you all had a probability of 4/52 for an ace. After the game has been played, probability will tell you that 10 aces in a row was unlikely to happen by chance.

As I have said before you are not looking at probability, you are looking at something more akin to predestination which cannot be described by probability.
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Offline alancalverd

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Re: Is this maths correct?
« Reply #128 on: 14/07/2015 23:29:58 »
You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card,

If you know this, the decks have probably not been shuffled fairly. Out of 1,000,000 decks the expected number with an ace as second card is 76,923, not 100,000. A 24% discrepancy over a million trials is good evidence of an irregularity.


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Offline Colin2B

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Re: Is this maths correct?
« Reply #129 on: 15/07/2015 13:39:10 »
You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card,

If you know this, the decks have probably not been shuffled fairly. Out of 1,000,000 decks the expected number with an ace as second card is 76,923, not 100,000. A 24% discrepancy over a million trials is good evidence of an irregularity.
It is also interesting to note that if TB understood probability he would also understand that even assuming we know the higher incidence of aces, deck skipping has no effect on the probability of receiving an ace.
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Offline Thebox

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Re: Is this maths correct?
« Reply #130 on: 15/07/2015 19:03:14 »



It is also interesting to note that if TB understood probability he would also understand that even assuming we know the higher incidence of aces, deck skipping has no effect on the probability of receiving an ace.

Thank you Colin and Alan for continuing with this conversation and at the same time giving me more knowledge.
Colin, you are truly starting to understand me, and you are working out what I mean in my posts.

''I think you imagine the decks being set out, with their shuffled order and by chance, unknown to all the players, 10% of the decks have aces 2nd card. This is a possible scenario,''


The reason I mentioned the second card is that for each hand, a player receives a different card position of the sequence relative to their seat position relative to the small blind, the small blind always gets the first card dealt, then the big blind gets the second card, and so up to 9 players.  Every hand the small blind moves around a seat clockwise.

''If you know this, the decks have probably not been shuffled fairly. Out of 1,000,000 decks the expected number with an ace as second card is 76,923, not 100,000. A 24% discrepancy over a million trials is good evidence of an irregularity.''


My example was just an example, 100,000 was just made up for my example.  but now you have gave me the correct values, thanks.

Colin mentions what actually happens, yes of cause that is what I am talking about like always with all my science, I always talk about what actually happens or what we actually observe and what we actually do know.
I was never insane, I just always see reality and the truths.

7
6
9
2
3

y axis


4

x axis


would you agree that in this scenario that x≠y?









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Offline alancalverd

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Re: Is this maths correct?
« Reply #131 on: 15/07/2015 23:57:24 »
It doesn't matter where you sit or when the cards are dealt. The probability of receiving an ace in any chair at any nominated point in any deal is 1/13.
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Offline Thebox

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Re: Is this maths correct?
« Reply #132 on: 16/07/2015 20:47:12 »
It doesn't matter where you sit or when the cards are dealt. The probability of receiving an ace in any chair at any nominated point in any deal is 1/13.

I asked a question, and that does not answer my question Alan. Can you please answer my question, in the scenario I provided I asked if x was not equal to y, 
y axis
7
6
9
2
3

x axis
4

x≠y would this equation be true for my provided scenario,

p.s y could have 13/13

« Last Edit: 16/07/2015 21:28:59 by Thebox »

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Offline alancalverd

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Re: Is this maths correct?
« Reply #133 on: 16/07/2015 23:41:26 »
None of the numbers in the set you call y is 4. So  what?

If you write down any finite set of integers, you can always find an integer that is not a member of that set. And you can also find a number that is. So what?
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Offline Thebox

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Re: Is this maths correct?
« Reply #134 on: 17/07/2015 08:06:20 »
None of the numbers in the set you call y is 4. So  what?

If you write down any finite set of integers, you can always find an integer that is not a member of that set. And you can also find a number that is. So what?


''Out of 1,000,000 decks the expected number with an ace as second card is 76,923''

7
6
9
2
3


4


You still avoided the actual question, can you not see the significant difference between x and y?

X is obviously not equal to y, I am obviously correct about online poker, I do not know what we are talking about, it may be probability or it may be what actual happens, but either way playing just x is not the same as playing y.  This is apparent.

We can not say that y is equal to x in any sense, y will have a certainty of repeat values, i.e ace of diamonds lets say 100 times where x can only have one.


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Offline alancalverd

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Re: Is this maths correct?
« Reply #135 on: 17/07/2015 08:50:59 »
You can't "play y". In poker you play one hand at a time. Obviously x is not equal to y because you are comparing apples and chickens.

There is no certainty in a fair shuffle, but because y is infinite and x is finite, the actual distribution in y is more likely to be close to the calculated probability than the actual distribution in any one x.

This kind of second-order statistics comes under the heading of confidence limits:the greater the number of trials, the greater the confidence we can have that the actual and expected distributions will converge.

However if all the trials are independent random shuffles, the expected distributions must be identical.
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Offline Thebox

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Re: Is this maths correct?
« Reply #136 on: 17/07/2015 09:06:23 »
You can't "play y". In poker you play one hand at a time. Obviously x is not equal to y because you are comparing apples and chickens.

There is no certainty in a fair shuffle, but because y is infinite and x is finite, the actual distribution in y is more likely to be close to the calculated probability than the actual distribution in any one x.

This kind of second-order statistics comes under the heading of confidence limits:the greater the number of trials, the greater the confidence we can have that the actual and expected distributions will converge.

However if all the trials are independent random shuffles, the expected distributions must be identical.

One hand at a time of the x axis and not of the y axis,

if you are picking a deck, you are playing the y axis, which is not equal to x.


123
123
123
123
123


we can clearly see this in simple diagram form, we firstly play the Y axis then play the x axis, that is why I called it cross odds or interception , the chance of receiving a deck that the second card is an ace using 1,000,000 decks, is obviously (76,923/1,000,000)/t= 0.076923/t and then  once the choice is made of deck  (4/52)/t=0.07692307692.This is all good as long as no other tables are involved.

(0.07692307692/t)(/n)/t  where n is table.

added - sorry that does not work,

P(x)=0.07692307692∩n
..................t





« Last Edit: 17/07/2015 09:19:50 by Thebox »

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Offline Colin2B

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Re: Is this maths correct?
« Reply #137 on: 17/07/2015 12:21:24 »
OK, Alan & I understand what you are trying to say, but as you say this is not standard probability but what you call cross-odds or interception.

If you are going to convince people you are right you need to think like a scientist and follow the scientific method. There are 2 things you need to do.

1. Get the maths right, otherwise people will just criticise your maths and not understand your ideas
2. You need to devise an experiment either using actual data from your games or via a large scale test so that your theory can be compared to what probability predicts.

First the maths:

the chance of receiving a deck that the second card is an ace using 1,000,000 decks, is obviously (76,923/1,000,000)/t= 0.076923/t and then  once the choice is made of deck  (4/52)/t=0.07692307692.This is all good as long as no other tables are involved.

(0.07692307692/t)(/n)/t  where n is table.

added - sorry that does not work,

P(x)=0.07692307692∩n
..................t

As I've said before you can't divide a probability by time, What do you divide by? 10 mins, 15s?
In your system time defines the move to the next deck. So you need to find the average game length, max & min times and devise a formula the will advance the deck selection to the next deck. So your formula would have to have a result something like =yi+1
Also people will pick holes in P(x)=0.07692307692∩n. If n is table number it equals a set of integers, this will not intersect with a probability, so you need a different way of expressing what you want to say.


Experimental Verification

Ideally you would take all the data provided by the gaming company and analyse it to compare with the predictions of probability theory and look at the differences.
If you don't have access to that you would need to either keep records as you play or devise an experiment to prove your theory.
You ought to be able to set up an experiment either using small packs of cards eg deck of 4 cards containing a heart, diamond, spade and club.
Or you could set up a spreadsheet to create random decks of cards and compare sequential vs deck skipping. You would need to have it go through a large number eg 1000.

What do you think? You could offer genuine scientific proof of your theory.

and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #138 on: 18/07/2015 08:06:57 »



What do you think? You could offer genuine scientific proof of your theory.

Thank you Colin, I also understand this is not standard and is an advancement on probabilities .   It is new and therefore the maths will be new.  I understand I must be explicit in explanation and maths, I do agree.

I know from observation I can observe the anomaly and witness the receiving of, example - 3 clubs 5/7 hands.  This happens regular, repeat values, I even have had the exact same hand from previous of both cards.

I go on holiday in the next hour so will be unable to reply for a week.

please see this,

http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&p=1391841#p1391841

#14844 page 594


a wave function difference of probability.

and I am not trying to explain divided by time, but over a period of time.


cross odds= P(n) from (x)≠P(n) from (y) in a period of time.

..yyy
x123x
x231x
x123x
..yyy

What else can we call this?

How else can we describe it?

And is it ''none standard deviance'' I am looking for?


added - in my link , x is constant over time where y is random over time.










« Last Edit: 18/07/2015 08:14:27 by Thebox »

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Offline Thebox

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Re: Is this maths correct?
« Reply #139 on: 24/07/2015 18:51:32 »
Hello , I am back, no comments for the last post?

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Offline alancalverd

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Re: Is this maths correct?
« Reply #140 on: 26/07/2015 12:49:22 »
OK, let's try this.

It's pretty clear that you want to be dealt a winning hand.

It also seems that you have an idea that, having entered the game and played a hand,  you can maximise your probability of being dealt a winning hand by not playing the next hand, but choosing another one.

So please complete the sentence "in order to maximise the probability of being dealt a winning hand on my second play, I should...."
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Offline Thebox

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Re: Is this maths correct?
« Reply #141 on: 27/07/2015 20:30:04 »
OK, let's try this.

It's pretty clear that you want to be dealt a winning hand.

It also seems that you have an idea that, having entered the game and played a hand,  you can maximise your probability of being dealt a winning hand by not playing the next hand, but choosing another one.

So please complete the sentence "in order to maximise the probability of being dealt a winning hand on my second play, I should...."

Hi Alan, completely different to what I have said. I do not want to be dealt a winning hand, I want to be dealt winning hands over x amount of time and not winning hands over y amount of time.
(note this time not axis's)


Spacing is important to the fundamental structure of Texas holdem poker.   Spacing of hands over a period of time/game is what makes losers or winners.
Players get huge stacks in minutes, then wait and wait and wait and nothing else comes. Of cause the vice versus, a player can receive no hands early on then a flourish of good hands in the later stages.
This does sometimes happen live as well, but there is differences online that destroy probabilities over a period of time/game.
I could get some hand history and show you, only the other night  I received the same card several times in only several hands, an often occurrence that happens.
(I can not repeat this with a real deck of cards)
How can I explain it simply, the order of distribution probabilities online is not the same as in a live game using one deck.  The spacing is different over time , sometimes there is no spacing,





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Offline alancalverd

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Re: Is this maths correct?
« Reply #142 on: 27/07/2015 21:59:50 »
The reason is simple. Live games do not employ truly random shuffles. Cards stick together to some extent, the dealer's fingers are not symmetrical, the cards are collected in groups....in fact in some games the cards are never shuffled but just cut so the hands continue to improve, whereas your beneficiaries have gone to great lengths to show how the online game is completely random.
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Offline Colin2B

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Re: Is this maths correct?
« Reply #143 on: 27/07/2015 23:23:31 »
The reason is simple. Live games do not employ truly random shuffles.
This has been studied by games researchers who discovered all sorts of no random patterns. To truly shuffle a pack requires a careful technique to ensure no bias.
Look back at my post where I talk about experimental verification. You need to have detailed records of your games and compare them to the expected results from probability theory.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #144 on: 28/07/2015 06:23:26 »
The reason is simple. Live games do not employ truly random shuffles. Cards stick together to some extent, the dealer's fingers are not symmetrical, the cards are collected in groups....in fact in some games the cards are never shuffled but just cut so the hands continue to improve, whereas your beneficiaries have gone to great lengths to show how the online game is completely random.

and truly random has no probabilities because it is infinite unlike a selective group of variants which is not infinite and a block of variants with a start and an end. 0 on a roulette wheel as got to come in time the same as an ace from a deck as to come in time. 

I think you still are missing the point, lets say in an average live game of 100 hands I receive a total of 5 aces in that period of time, the aces are spaced out by other hands.  Online it is possible to receive 100 aces in 100 hands from using a Y axis.



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Offline Colin2B

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Re: Is this maths correct?
« Reply #145 on: 28/07/2015 10:09:38 »
truly random has no probabilities
No, there are still probabilities that can be calculated. Anyway, random what?

Online it is possible to receive 100 aces in 100 hands from using a Y axis.
Possible is not the same as probable.

and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #146 on: 28/07/2015 18:00:15 »
truly random has no probabilities
No, there are still probabilities that can be calculated. Anyway, random what?

Online it is possible to receive 100 aces in 100 hands from using a Y axis.
Possible is not the same as probable.

Likely and unlikely spring to mind, it is likely after receiving an ace in a live game, that the next hand you will be unlikely to get an ace although not impossible.

It is just as likely from an internet game that the next card you receive is an ace from a multitude of decks and more than a possible 1 of 4 aces, been in the first card position of multiple decks. i.e ace of diamonds 2000 times as the first card.

if there were 100000 decks and 2000 top cards were the ace of diamonds, I have a 2000/100000 chance of receiving a top card of the ace of diamonds, I do not believe that is 1/52

I believe this is  basic maths and undeniably obvious.

x=1/52
y=2000/100000

they are obviously not equal.





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Offline alancalverd

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Re: Is this maths correct?
« Reply #147 on: 28/07/2015 20:16:37 »

Likely and unlikely spring to mind, it is likely after receiving an ace in a live game, that the next hand you will be unlikely to get an ace although not impossible.


1 card in 13 is an ace. If you are dealt a hand of 5 cards the probability of getting 1 ace is therefore 5/13 in every hand you are dealt, if the pack is fully shuffled between hands. However if the pack is merely collected and cut, because nobody will discard the aces, so the collected hands will have more adjacent aces and

the probability of getting one ace in the next hand is less than 5/13 but

the probability of getting two aces in the next hand is greater than (5/13)^2.

But the fact remains that if the cards are fairly shuffled between deals, every hand has the same likelihood of any chosen distribution.


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Offline Colin2B

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Re: Is this maths correct?
« Reply #148 on: 30/07/2015 11:32:10 »
x=1/52
y=2000/100000

they are obviously not equal.
They are not equal because 2000/100000 is a number you have made up, it is not the true probability in the y direction. We have told you before what it is.
Anyway, this is not the crux of your problem, it is the issue of deck skipping and it's effect on probability.

Come on old friend, we can't keep going around in circles , you need to provide actual game logs to show whether there is a fault in the RNG, or distribution. The problem does n't lie with probability.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline alancalverd

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Re: Is this maths correct?
« Reply #149 on: 30/07/2015 16:53:57 »
Quote
if there were 100000 decks and 2000 top cards were the ace of diamonds, I have a 2000/100000 chance of receiving a top card of the ace of diamonds, I do not believe that is 1/52

If there were 100,000 fairly shuffled decks, the expectation is that top card would be the A◊ in 100,000/52 = 1923 cases. 1800 to 2000 cases in an actual 100,000 is not outside the realms of reasonable probability but I would suspect foul play at 1400 or 2400.
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